# Introduction

Given a z-shaped telescope setup with two beamsplitters such as depicted in Fig. 1, we calculate the effects on the outgoing beam when applying a small additional tilt β to the angle of incidence α of both beamsplitters.

Such a small extra tilt is applied in Finesse 2 via the xbeta parameter of a bs component. With respect to the “untilted beam” (i.e. β = 0) we calculate – for the outgoing beam – both the shift in its position and the change of its phase δφ, each as a function of the angle β and the distance s between the beamsplitters. For the shift we derive eq. (5) below

\Delta = s \cdot \sin 2\beta

while for the phase change we find eq. (13) below

\delta \varphi = \frac{2 s \cdot \sin^2 \beta}{\lambda} \cdot 360\degree

Note that this is the first post in a series, the second post (part 2: finesse examples) will compare the results obtained here with simulations from finesse 2.

# Detailed Setup

Fig. 2 depicts the two different trajectories. The distance between the two beamsplitters is s. The initial angle of incidence of the “untilted” beamsplitters (dark bue) is α and its corresponding trajectory is plotted in dark red. The two beamsplitters are rotated over a small additional angle β (light blue) with corresponding trajectory in light red.
As a result, the path traveled by the beam between the first and second beamsplitter changes from s to ; it will hit the second mirror at a different point (a distance d away from the original point) and the outgoing beam will shift over a distance Δ. The difference in total path length between the dark and light red paths will cause a phase shift δφ. In the next subsections we will derive Δ and followed by δφ.

## Calculation of shift in position Δ

Using the sine rule for the triangle enclosed by d, s and

\frac{d}{\sin 2\beta} = \frac{s}{\sin \gamma} = \frac{\tilde{s}}{\sin(90\degree+\alpha-\beta)} \qquad (1)

we can get the length d

d = s \cdot \frac{\sin 2\beta}{\sin \gamma} \qquad (2)

Furthermore from the total angle of that same triangle we can get the angle γ

(90\degree-\alpha-\beta+2\alpha) + 2\beta + \gamma = 180\degree \quad \to \quad \gamma = 90\degree – \alpha – \beta \qquad (3)

We need to calculate Δ:

\Delta = d \cdot \sin (90\degree-\alpha-\beta) = d \cdot \sin \gamma \qquad (4)

where we used eq. (3). Using eq. (2) we then obtain the shift

\Delta = s \cdot \sin 2\beta \qquad (5)

We see that all α dependence cancels out. Furthermore, we see that the shift increases linearly with the distance between the two beamsplitters. Since for larger values of β the paraxial approximation requires more and more higher modes to be included in a simulation, increasing s instead of β could help simulating the same shift Δ (we will look into convergence in the second post of this series).

## Calculation of difference in path length and δφ

Note that there are two parts:

1. p1: the difference between and s
2. p2: horizontal part since we hit the beamsplitter further to the right

p1 will have a positive contribution, while p2 will have a negative contribution to the change in path length.

### calculation p1

First note that

\sin \gamma = \sin(90\degree – \alpha – \beta) = \cos(\alpha + \beta) \qquad (6)

We can obtain from the sine rule eq. (1) and use (6) to get

\tilde{s} = s \cdot \frac{\sin(90\degree + \alpha – \beta)}{\sin \gamma} = s \cdot \frac{\cos(\alpha – \beta)}{\cos(\alpha + \beta)} \qquad (7)

so

p_1 = \tilde{s} – s = s \cdot \frac{\cos(\alpha – \beta) – \cos(\alpha + \beta)}{\cos(\alpha + \beta)} \qquad (8)

### calculation p2

Since part p2 shortens the total path, we add an overall minus sign:

p_2 = – d \cdot \cos(90\degree-\alpha-\beta) = -d \cdot \sin(\alpha + \beta) \qquad (9)

Using eq. (2) and (6) this becomes

p_2 = – s \cdot \frac{\sin (\alpha + \beta) \sin 2\beta}{\cos(\alpha + \beta)} \qquad (10)

### combining the two

Combining eq. (8) and (10) we get

s \cdot \frac{\cos (\alpha – \beta) – \cos (\alpha + \beta) – \sin (\alpha + \beta) \sin 2\beta}{\cos (\alpha + \beta)} \qquad (11)

We work out the numerator:

\cos (\alpha – \beta) – \cos (\alpha + \beta) – \sin (\alpha + \beta) \sin 2\beta =
\cos\alpha \cos\beta + \sin\alpha \sin\beta – \cos\alpha \cos\beta + \sin\alpha \sin\beta – (\sin\alpha \cos\beta + \cos\alpha \sin\beta) 2 \sin\beta \cos\beta =
2 \sin\alpha \sin\beta -2 (\sin\alpha \sin\beta \cos^2\beta + \cos\alpha \cos\beta \sin^2\beta) =
2 \sin\alpha \sin\beta ( 1 – \cos^2\beta) – 2 \cos\alpha \cos\beta \sin^2\beta =
2 (\sin\alpha \sin\beta – \cos\alpha \cos\beta) \sin^2\beta =
-2 \cos(\alpha + \beta) \sin^2 \beta

Putting this into eq. (11) we see that the cos(α+β) factor cancels out and we find the total path difference

p_1 + p_2 = -2 s \cdot \sin^2 \beta \qquad (12)

Note that again the α dependence cancels out. The phase shift resulting from the path difference can be calculated by dividing eq. (12) by the wavelength λ and multiplying by 360°. We also pick up an extra overall minus sign, cancelling the one from eq. (12), since a plane wave is defined as cos(ω tk x + φ), see for example Bond et al. (2016) eq. (1.4). Putting it together we get

\delta \varphi = \frac{2 s \cdot \sin^2 \beta}{\lambda} \cdot 360\degree \qquad (13)

## Conclusions

• Both the shift of the beam eq. (5) and the change in phase eq. (13) are independent of the angle α
• Given the square in eq. (12) the path length difference is small but not per se compared to the wavelength λ, leading to a significant phase shift.

# Effect finite size beam

The above calculation was done for an infinitesimally narrow beam hitting the beamsplitter exactly in the midpoint (its rotation point). We show here that the result is the same for a beam hitting the beamsplitter off-axis.